Convex set with no interior contained in hyperplane?

Question

Let $K$ be a convex set in a normed space $X$. Assume that $int(K)=\emptyset$ (norm topology). Must $K$ be contained in some (affine) hyperplane? It's fairly easy to see that this is true in $ℝ^n$, but i couldn't generalize.

Answer 1

Based on Jack's comment.

The "Hilbert cube" in Hilbert space $l^2$. $$C :=\{(x_1,x_2,\dots) : |x_k| \le 2^{-k}\;\forall k\}$$ $C$ is convex, compact (so it has empty interior) but has dense span (so it not contained in a closed hyperplane).

---

However, $C$ is contained in a (non-closed) hyperplane. (Axiom of Choice required.)
The linear span of $C$ is not the whole of $l^2$. Indeed, if $(a_1,a_2,\dots)$ is in the span of $C$, then $\limsup_k 2^k |a_k| < \infty$, but $(1,1/2,1/3,\cdots) \in l^2$ fails that property.
Now we define a Hamel basis for $l^2$ as follows: first choose any vector $u$ not in the span of $C$; next add to {u} a Hamel basis of the span of $C$; then extend that to a Hamel basis $B$ of $l^2$. Then we can see that $C$ is contained in the hyperplane $$ \left\{\sum_{b \in B} t_b b : t_u = 0\right\} $$

Answer 2

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$ with $\inf_{i\in I}\|v_i\|=0$. Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-1,1]\forall i\in I\right\}$. It has empty interior because for all $v\in X$ and all $i$, $v+3v_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.

Answer 3

Let $T\colon\mathcal{X}\to\mathcal{X}$ be a compact operator with a dense range on an infinite-dimensional separable Banach space $\mathcal{X}$ $.^{*}$ Then $K:=\overline{T(B_1(0))}$, where $B_1(0)$ is the unit ball, is a convex set with an empty interior (otherwise, ${\rm{span} }K=\mathcal{X}$, which contradicts the infinite-dimensionality of $\mathcal{X}$ because $K$ is compact). But via the dense range of $T$, we have $$ \mathcal{X}=\overline{\rm{span}} K=\overline{\rm{aff}}K \,, $$ and therefore $K$ cannot lie in any closed hyperplane $.^{**}$

$.^*$ Such operators always exist on (infinite-dimensional) separable Banach spaces.

$.^{**}$ $\rm{span}$ and $\rm{aff}$ denote, respectively, the (not necessarily closed) linear span and affine hull; for the latter case, allow complex scalars when $\mathcal{X}$ is over $\mathbb{C}$, otherwise simply assume $\mathcal{X}$ is over $\mathbb{R}$.

Answer 4

Here is a trick, modeled on Saúl RM's answer, to create huge convex sets without interior in an infinite-dimensional TVS.

Let $X$ be an infinite-dimensional TVS over $\mathbf{R}$. Take a Hamel basis $\{e_i\}_{i\in I}$ for $X$ and fix a total order on $I$ without a greatest element. (Such an order always exists, see here.) Then every $x\in X$ can be uniquely written as a finite sum $x = \lambda_1 e_{i_1} +\cdots +\lambda_n e_{i_n}$, where $i_1< i_2<\cdots<i_n$, $\lambda_i\neq 0$. Let $G$ be the set consisting of those $x\in X$ such that $\lambda_n > 0$. Then $G$ is clearly convex, $G\cap (-G) =\varnothing$, $G\cup (-G) = X\backslash \{0\}$. (So $G$ is huge, thus not contained in any hyperplane, closed or not.)

Claim: $\mathbb{int}\, G=\varnothing$. Indeed, let $x = \lambda_1 e_{i_1} +\cdots +\lambda_n e_{i_n}\in G$, and let $U$ be a neighborhood of $x$. Choose an $e_i$ such that $i> i_n$. Then $U-x$ is a neighborhood of $0$. Choose $\delta>0$ so small such that $-\delta e_i \in U-x$. Then $-\delta e_i+x\in U$ but $-\delta e_i+x\notin G$.